Study Guide and Solutions Manual

Organic Chemistry
A Brief Course
THIRTEENTH EDITION

David Hart
Ohio State University

Chris Hadad
Ohio State University

David Craine
Ohio State University

Harold Hart

Michigan State University

Prepared by
David Brown
Florida Gulf Coast University

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Contents
Introduction to the Student..................................................................................................v
Chapter 1: Bonding and Isomerism .....................................................................................1
Chapter 2: Alkanes and Cycloalkanes; Conformational and Geometric Isomerism .........18
Chapter 3: Alkenes and Alkynes........................................................................................34
Chapter 4: Aromatic Compounds ......................................................................................58
Chapter 5: Stereoisomerism ............................................................................................. 82
Chapter 6: Organic Halogen Compounds; Substitution and Elimination Reactions .......102
Chapter 7: Alcohols, Phenols, and Thiols........................................................................116
Chapter 8: Ethers and Epoxides.......................................................................................134
Chapter 9: Aldehydes and Ketones..................................................................................149
Chapter 10: Carboxylic Acids and Their Derivatives......................................................175
Chapter 11: Amines and Related Nitrogen Compounds..................................................198
Chapter 12: Spectroscopy and Structure Determination..................................................219
Chapter 13: Heterocyclic Compounds .............................................................................233
Chapter 14: Synthetic Polymers.......................................................................................249
Chapter 15: Lipids and Detergents ..................................................................................263
Chapter 16: Carbohydrates ..............................................................................................274
Chapter 17: Amino Acids, Peptides, and Proteins...........................................................298

Chapter 18: Nucleotides and Nucleic Acids ....................................................................326
Summary of Synthetic Methods...............................................................................341
Summary of Reaction Mechanisms .........................................................................354
Review Problems On Synthesis ...............................................................................359
Sample Multiple Choice Questions .........................................................................363
iii
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Introduction to the Student
This study guide and solutions book was written to help you learn organic chemistry. The
principles and facts of this subject are not easily learned by simply reading them, even
repeatedly. Formulas, equations, and molecular structures are best mastered by written
practice. To help you become thoroughly familiar with the material, we have included many
problems within and at the end of each chapter in the text.
It is our experience that such questions are not put to their best use unless correct
answers are also available. Indeed, answers alone are not enough. If you know how to work
a problem and find that your answer agrees with the correct one, fine. But what if you work
conscientiously, yet cannot solve the problem? You then give in to temptation, look up the
answer, and encounter yet another dilemma–how in the world did the author get that
answer? This solutions book has been written with this difficulty in mind. For many of the
problems, all of the reasoning involved in getting the correct answer is spelled out in detail.
Many of the answers also include cross-references to the text. If you cannot solve a
particular problem, these references will guide you to parts of the text that you should
review.

Each chapter of the text is briefly summarized. Whenever pertinent, the chapter
summary is followed by a list of all the new reactions and mechanisms encountered in that
chapter. These lists should be especially helpful to you as you review for examinations.
When you study a new subject, it is always useful to know what is expected. To help
you, we have included in this study guide a list of learning objectives for each chapter—that
is, a list of what you should be able to do after you have read and studied that chapter. Your
instructor may want to delete items from these lists of objectives or add to them. However,
we believe that if you have mastered these objectives—and the problems should help you to
do this—you should have no difficulty with examinations. Furthermore, you should be very
well prepared for further courses that require this course as a prerequisite.
Near the end of this study guide you will find additional sections that may help you to
study for the final examination in the course. The SUMMARY OF SYNTHETIC METHODS
lists the important ways to synthesize each class of compounds discussed in the text. It is
followed by the SUMMARY OF REACTION MECHANISMS. Both of these sections have
references to appropriate portions of the text, in case you feel that further review is
necessary. Finally, you will find two lists of sample test questions. The first deals with
synthesis, and the second is a list of multiple-choice questions. Both of these sets should
help you prepare for examinations.
In addition, we offer you a brief word of advice about how to learn the many
reactions you will study during this course. First, learn the nomenclature systems thoroughly
for each new class of compounds that is introduced. Then, rather than memorizing the
particular examples of reactions given in the text, study reactions as being typical of a class
of compounds. For example, if you are asked how compound A will react with compound B,
proceed in the following way. First ask yourself: to what class of compounds does A belong?
How does this class of compounds react with B (or with compounds of the general class to
which B belongs)? Then proceed from the general reaction to the specific case at hand. This
approach will probably help you to eliminate some of the memory work often associated with
organic chemistry courses. We urge you to study regularly, and hope that this study guide
and solutions book will make it easier for you to do so.


v
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vi

Introduction to the Student

Great effort has been expended to ensure the accuracy of the answers in this book
and we wish to acknowledge the helpful comments provided by David Ball (Cleveland State
University) in this regard.

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1
Bonding and Isomerism
Chapter Summary
An atom consists of a nucleus surrounded by electrons arranged in orbitals. The electrons
in the outer shell, or the valence electrons, are involved in bonding. Ionic bonds are
formed by electron transfer from an electropositive atom to an electronegative atom.
Atoms with similar electronegativities form covalent bonds by sharing electrons. A single
bond is the sharing of one electron pair between two atoms. A covalent bond has specific
bond length and bond energy.
Carbon, with four valence electrons, mainly forms covalent bonds. It usually forms
four such bonds, and these may be with itself or with other atoms such as hydrogen,

oxygen, nitrogen, chlorine, and sulfur. In pure covalent bonds, electrons are shared equally,
but in polar covalent bonds, the electrons are displaced toward the more electronegative
element. Multiple bonds consist of two or three electron pairs shared between atoms.
Structural (or constitutional) isomers are compounds with the same molecular
formulas but different structural formulas (that is, different arrangements of the atoms in
the molecule). Isomerism is especially important in organic chemistry because of the
capacity of carbon atoms to be arranged in so many different ways: continuous chains,
branched chains, and rings. Structural formulas can be written so that every bond is shown,
or in various abbreviated forms. For example, the formula for n-pentane (n stands for
normal) can be written as:
H

H

H

H

H

H

C

C

C

C


C

H

H

H

H

H

H

or

CH3CH2 CH2 CH2CH3

or

Some atoms, even in covalent compounds, carry a formal charge, defined as the
number of valence electrons in the neutral atom minus the sum of the number of unshared
electrons and half the number of shared electrons. Resonance occurs when we can write
two or more structures for a molecule or ion with the same arrangement of atoms but
different arrangements of the electrons. The correct structure of the molecule or ion is a
resonance hybrid of the contributing structures, which are drawn with a double-headed
arrow ( ) between them. Organic chemists use a curved arrow ( ) to show the movement
of an electron pair.
A sigma ( ) bond is formed between atoms by the overlap of two atomic orbitals
along the line that connects the atoms. Carbon uses sp3-hybridized orbitals to form four

such bonds. These bonds are directed from the carbon nucleus toward the corners of a
tetrahedron. In methane, for example, the carbon is at the center and the four hydrogens
are at the corners of a regular tetrahedron with H–C–H bond angles of 109.5°.
In the chapter summaries, terms whose meanings you should know appear in boldface type.

1
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2

Chapter 1

Carbon compounds can be classified according to their molecular framework as
acyclic (not cyclic), carbocyclic (containing rings of carbon atoms), or heterocyclic
(containing at least one ring atom that is not carbon). They may also be classified according
to functional group (Table 1.6).

Learning Objectives
1.

Know the meaning of: nucleus, electrons, protons, neutrons, atomic number, atomic
weight, shells, orbitals, valence electrons, valence, kernel.

2.

Know the meaning of: electropositive, electronegative, ionic and covalent bonds,
radical, catenation, polar covalent bond, single and multiple bonds, nonbonding or
unshared electron pair, bond length, bond energy.


3.

Know the meaning of: molecular formula, structural formula, structural (or
constitutional) isomers, continuous and branched chain, formal charge, resonance,
contributing structures, sigma ( ) bond, sp3-hybrid orbitals, tetrahedral carbon.

4.

Know the meaning of: acyclic, carbocyclic, heterocyclic, functional group.

5.

Given a periodic table, determine the number of valence electrons of an element and
write its electron-dot formula.

6.

Know the meaning of the following symbols:
+

–

7.

Given two elements and a periodic table, tell which element is more electropositive
or electronegative.

8.


Given the formula of a compound and a periodic table, classify the compound as
ionic or covalent.

9.

Given an abbreviated structural formula of a compound, write its electron-dot
formula.

10.

Given a covalent bond, tell whether it is polar. If it is, predict the direction of bond
polarity from the electronegativities of the atoms.

11.

Given a molecular formula, draw the structural formulas for all possible structural
isomers.

12.

Given a structural formula abbreviated on one line of type, write the complete
structure and clearly show the arrangement of atoms in the molecule.

13.

(pentane), write the complete structure and
Given a line formula, such as
clearly show the arrangement of atoms in the molecule. Tell how many hydrogens
are attached to each carbon, what the molecular formula is, and what the functional
groups are.


14.

Given a simple molecular formula, draw the electron-dot formula and determine
whether each atom in the structure carries a formal charge.

Although the objectives are often worded in the form of imperatives (i.e., determine …,write …, draw …), these
verbs are all to be preceded by the phrase “be able to …”. This phrase has been omitted to avoid repetition.

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Bonding and Isomerism

3

15.

Draw the electron-dot formulas that show all important contributors to a resonance
hybrid and show their electronic relationship using curved arrows.

16.

Predict the geometry of bonds around an atom, knowing the electron distribution in
the orbitals.

17.

Draw in three dimensions, with solid, wedged, and dashed bonds, the tetrahedral

bonding around sp3-hybridized carbon atoms.

18.

Distinguish between acyclic, carbocyclic, and heterocyclic structures.

19.

Given a series of structural formulas, recognize compounds that belong to the same
class (same functional group).

20.

Begin to recognize the important functional groups: alkene, alkyne, alcohol, ether,
aldehyde, ketone, carboxylic acid, ester, amine, nitrile, amide, thiol, and thioether.

ANSWERS TO PROBLEMS
Problems Within the Chapter
1.1

The lithium atom donates its valence electron to the bromine atom to form the ionic
compound, lithium bromide.
Li

+

Br

Li + +


Br

–

1.2

Elements with fewer than four valence electrons tend to give them up and form
positive ions: Al3+, Li+. Elements with more than four valence electrons tend to gain
electrons to complete the valence shell, becoming negative ions: S2–, O2–.

1.3

Within any horizontal row in the periodic table, the most electropositive element
appears farthest to the left. Na is more electropositive than Al, and B is more
electropositive than C. In a given column in the periodic table, the lower the element,
the more electropositive it is. Al is more electropositive than B.

1.4

Within any horizontal row in the periodic table, the most electronegative element
appears farthest to the right. F is more electronegative than O, and O more than N.
In a given column, the higher the element, the more electronegative it is. F is more
electronegative than Cl.

1.5

As will be explained in Sec. 1.3, carbon is in Group IV and has a half-filled (or halfempty) valence shell. It is neither strongly electropositive nor strongly
electronegative.

1.6


The unpaired electrons in the fluorine atoms are shared in the fluorine molecule.
F + F
fluorine atoms

1.7

dichloromethane (methylene chloride)
H
H C Cl
Cl

1.8

H

or H

C Cl
Cl

F F + heat
fluorine molecule

trichloromethane (chloroform)
Cl
H C Cl
Cl

Cl


or H

C Cl
Cl

If the C–C bond length is 1.54 Å and the Cl–Cl bond length is 1.98 Å, we expect the
C–Cl bond length to be about 1.76 Å: (1.54 + 1.98)/2. In fact, the C–Cl bond (1.75 Å)
is longer than the C–C bond.

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4

Chapter 1

1.9

Propane
H

H

H

H

C


C

C

H

H

H

H

1.10

N +–Cl – ; S +–O – The key to predicting bond polarities is to determine the relative
electronegativities of the elements in the bond. In Table 1.4, Cl is more
electronegative than N. The polarity of the S–O bond is easy to predict because both
elements are in the same column of the periodic table, and the more electronegative
atom appears nearer the top.

1.11

Both Cl and F are more electronegative than C.
–

F
–

+


Cl C Cl

–

F–

1.12

Both the C–O and H–O bonds are polar, and the oxygen is more electronegative
than either carbon or hydrogen.
H

H

C

O

H

H

1.13
1.14

H C

a.
b.

c.

1.15

N

H C

N

H C

N

The carbon shown has 12 electrons around it, 4 more than are allowed.
There are 20 valence electrons shown, whereas there should only be 16 (6
from each oxygen and 4 from the carbon).
There is nothing wrong with this formula, but it does place a formal charge of
–1 on the “left” oxygen and +1 on the “right” oxygen (see Sec. 1.11). This
formula is one possible contributor to the resonance hybrid structure for
carbon dioxide (see Sec. 1.12); it is less important than the structure with two
carbon–oxygen double bonds, because it takes energy to separate the + and
– charges.

Methanal (formaldehyde), H2CO. There are 12 valence electrons altogether (C = 4,
H = 1, and O = 6). A double bond between C and O is necessary to put 8 electrons
around each of these atoms.
H
H C


H
O

or

C

O

H

1.16

There are 10 valence electrons, 4 from C and 6 from O. An arrangement that puts
8 electrons around each atom is shown below. This structure puts a formal charge of
–1 on C and +1 on O (see Sec. 1.11).
C

O

or

C

O

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Bonding and Isomerism

1.17

5

If the carbon chain is linear, there are two possibilities:
H

H

C

C

C

H

H

H

H
C

H

H


H

and

H

H

H

H

C

C

C

C

H

H

H

But the carbon chain can be branched, giving a third possibility:
H
H


C
C

C

H

C
H

1.18

a.
H

H

H

C

N

H

H

H
H
H

H

b.

H

H

C

O

H

H

Notice that the nitrogen has one non-bonded electron pair (part a) and the oxygen
has two non-bonded electron pairs (part b).
1.19

No, it does not. We cannot draw any structure for C2H5 that has four bonds to each
carbon and one bond to each hydrogen.

1.20

First write the alcohols (compounds
with an O–H group).

H


H

H

H

C

C

C

H

H

H

O

H

and

H

H

H


H

C

C

C

H

O

H

H

H

Then write the structures with a
C–O–C bond (ethers).

H

H

C

O

H


There are no other possibilities. For
example,

H

O

H

H

H

C

C

C

H

H

H

H

H


H

C

C

H

H

and

H

H
H

H

H

C

C

C

H

H


O

H

H

are the same as
H

H

H

H

C

C

C

H

H

H

O


H

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6

Chapter 1

They all have the same bond connectivities and represent a single structure. Similarly,
H
H

C

O

H

H

H

C

C

H


H

H

is the same as

H

H

H

C

C

H

H

1.21

From left to right: n-pentane, isopentane, and isopentane.

1.22

a.

H
H

H C
H C
H C
H
H

b.

H

Cl

O

H

O

C

H

H

Cl
C

C

H


C

Cl

Cl

H

Notice that the non-bonded electron pairs on oxygen and chlorine are not shown.
Non-bonded electron pairs are frequently omitted from organic structures, but it is
important to know that they are there.0
1.23

First draw the carbon skeleton showing all bonds between carbons.
C

C

C

C

C

C

Then add hydrogens to satisfy the valency of four at each carbon.

H

H

H

C

H

H

H

C

C

H
H
H

C
H

H

C

C

H


H

CH2

or

H3C

CH
CH2

H

stands for the carbon skeleton

C

C
C

CH3

C

C

1.24

CH3


C

C

Addition of the appropriate number of hydrogens on each carbon completes the
valence of 4.
H

1.25

ammonia

H

formal charge on nitrogen = 5 – (2 + 3) = 0

N

H

ammonium ion

+

H
H

N


H

formal charge on nitrogen = 5 – (0 + 4) = +1

H

amide ion

H
N

–

formal charge on nitrogen = 5 – (4 + 2) = –1

H

The formal charge on hydrogen in all three cases is zero [1 – (0 + 1) = 0].

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Bonding and Isomerism

1.26

7

For the singly bonded oxygens, formal charge = 6 – (6 + 1) = –1.

For the doubly bonded oxygen, formal charge = 6 – (4 + 2) = 0.
For the carbon, formal charge = 4 – (0 + 4) = 0.
2–

O
O

1.27

C

O

There are 24 valence electrons to use in bonding (6 from each oxygen, 5 from the
nitrogen, and one more because of the negative charge). To arrange the atoms with
8 valence electrons around each atom, we must have one nitrogen–oxygen double
bond:
O
O

N

O
O

O

N

–


O
O

O

N

O

The formal charge on nitrogen is 5 – (0 + 4) = +1.
The formal charge on singly bonded oxygen is 6 – (6 + 1) = –1.
The formal charge on doubly bonded oxygen is 6 – (4 + 2) = 0.
The net charge of the ion is –1 because each resonance structure has one positively
charged nitrogen atom and two negatively charged oxygen atoms. In the resonance
hybrid, the formal charge on the nitrogen is +1; on the oxygens, the charge is –2/3 at
each oxygen, because each oxygen has a –1 charge in two of the three structures
and a zero charge in the third structure.
1.28

There are 16 valence electrons (five from each N plus one for the negative charge).
The formal charges on each nitrogen are shown below the structures.
N

N

N

–1 +1 –1


N

N

N

0 +1 –2

1.29

In tetrahedral methane, the H–C–H bond angle is 109.5°. In “planar” methane, this
angle would be 90o and bonding electrons would be closer together. Thus, repulsion
between electrons in different bonds would be greater in “planar” methane than in
tetrahedral methane. Consequently, “planar” methane would be less stable than
tetrahedral methane.

1.30

a.
b.
c.
d.

C=O, ketone; C=C, alkene; O–H, alcohol
arene; C(=O)NH, amide; C–S–C, thioether, C(=O)O–H, carboxylic acid
C=O, ketone
C=C, alkene

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8

Chapter 1

ADDITIONAL PROBLEMS
1.31

The number of valence electrons is the same as the number of the group to which
the element belongs in the periodic table (Table 1.3).
b.
c.
d.
a.
Ca
O
C
B
e.

f.

P

covalent
ionic

Br


1.32

a.
e.

1.33

The bonds in sodium chloride are ionic; Cl is present as chloride ion (Cl–); Cl– reacts
with Ag+ to give AgCl, a white precipitate. The C–Cl bonds in CCl4 are covalent, no
Cl– is present to react with Ag+.

1.34
a.
b.
c.
d.
e.
f.

b.
f.

ionic
covalent

Valence Electrons
6
1
6
4

5
7

O
H
S
C
N
Cl

c.
g.

covalent
ionic

d.
h.

covalent
covalent

Common Valence
2
1
2
4
3
1


Note that the sum of the number of valence electrons and the common valence is 8
in each case. (The only exception is H, where it is 2, the number of electrons in the
completed first shell.)
1.35

a.
H

d.
H

1.36

a.

H

H

C

C

H

H

H

H


C

N

H

H

+

b.
O

H

c.

H

H

C

F

H

H


e.
H

H

H

C

C Cl

H

H

f.

H

H

H

C

C

C

H


H

H

H

H
C

O

H

Bromine is more electronegative than hydrogen.

–

H Br

b.

H

H

-

+


C

Fluorine is more electronegative than carbon.

F

H

c.
d.

–

+

–

O

C

O

Cl Cl

The C=O bond is polar, and the oxygen is more
electronegative than carbon.

Since the bond is between identical atoms, it is
pure covalent (nonpolar).


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Bonding and Isomerism

e.

9

Fluorine is more electronegative than sulfur.
Indeed, it is the most electronegative element.
Note that the S has 12 electrons around it.
Elements below the first full row sometimes have
more than 8 valence electrons around them.

-

F

F
F

+

-

F
-


f.

S

F

F

-

-

Carbon and hydrogen have nearly identical
electronegativities, and the bonds are essentially
nonpolar.

H
H

C

H

H

g.

–


O

+

S

O

H

h.
H

C

Oxygen is more electronegative than sulfur.

–

+

O

Oxygen is more electronegative than carbon, or
hydrogen, so the C–O and O–H bonds are polar
covalent.

+

H


H

1.37

The O–H bond is polar (there is a big difference in electronegativity between oxygen and
hydrogen), with the hydrogen +. The C–H bonds in acetic acid are not polar (there is
little electronegativity difference between carbon and hydrogen). The negatively charged
oxygen of the carbonate hydroxide deprotonates the acetic acid.
O

O
+ –

C
H3 C

1.38

a.

+ Na

OH

OH

C3H6

C

H3 C

O– Na+

+ H2 O

There must be a double bond or, with three carbons, a ring:
CH3 CH

CH2

or

CH2
H2 C

CH2

b.

C3H7Cl

The chlorine can replace a hydrogen on an end carbon or on
the middle carbon in the answer to a: CH3CH2CH2Cl or
CH3CH(Cl)CH3

c.

C2H4F2


The fluorines can either be attached to the same carbon or to
different carbons:
CH3CHF2 or CH2FCH2F

d.

C3H8

The only possible structure is CH3CH2CH3.

e.

C4H9F

The carbon chain may be either linear or branched. In each
case there are two possible positions for the fluorine.
CH3CH2CH2CH2F
(CH3)2CHCH2F

f.

C3H6Cl2

CH3CH(F)CH2CH3
(CH3)3CF

Be systematic. With one chlorine on an end carbon, there are
three possibilities for the second chlorine.

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10

Chapter 1

H

Cl H

H

C

C

C

Cl H

H

H Cl H

H

C

H


C

C

Cl H

H

H

H

H

H Cl

C

C

C

Cl H

H

H

If one chlorine is on the middle atom, the only new structure

arises with the second chlorine also on the middle carbon:
H Cl H
H

C

C

C

H

H Cl H

g.

C4H10O

With an O–H bond, there are four possibilities:
CH3CH2CH2CH2–OH
(CH3)2CHCH2–OH

CH3CH(OH)CH2CH3
(CH3)3C–OH

There are also three possibilities in a C–O–C arrangement:
CH3–O–CH2CH2CH3
CH3–O–CH(CH3)2
h.


1.39

C2H2I 2

CH3CH2–O–CH2CH3

The sum of the hydrogens and iodines is 4, not 6. Therefore
there must be a carbon–carbon double bond: I2C=CH2 or
CHI=CHI. No carbocyclic structure is possible because there
are only two carbon atoms.

The problem can be approached systematically. Consider first a chain of six carbons,
then a chain of five carbons with a one-carbon branch, and so on.
CH3

CH2

CH2

CH2

CH2

CH3

CH3

CH2

CH2


CH

CH3

CH3
CH3

CH2

CH

CH2

CH3

CH3
CH3

CH

CH

CH3

CH3

CH3

CH2


C

CH3 CH3

1.40

a.
H

CH3

CH3

H

H

H

H

H

H

C

C


C

C

C

C

H

H

H

H

H

H

b.

H

H

H

H
H


C

C
H
H

H
H

H

H

C

C

C

C

H

H

C

H
H


H

H

C

C

H

H

H

H

c.
H

H

H

H

H

H


C

C

N

C

C

H

H

H

H

d.
H

H

H

H

C

C


H

H

S

H

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Bonding and Isomerism

e.
H

11

H

H

C

C

H


f.
H

Cl O

H

H

1.41

a.

CH3 CH2C(CH3 )

H

H

O

H

H

C

C

C


C

C

H

H

H

H

H

b.

CH2

H

H
H
C

C

H
C


C
H

H

C
H

c.

d.

(CH3)3CCH(CH3)2

CH3CH2CH2CHCH2CHCH2CH2CH3
CH3

e.
g.

H
H
H

H
H

a.

(CH3)2CHCH2CH2CH 3


h.

H

H2C

H

C
C

C

C

C
C

H

1.42

f.

CH3CH2OCH 2CH3

CH2

H2C


CH2
O

H
H

In line formulas, each line segment represents a C–C bond, and C–H bonds
are not shown (see Sec. 1.10). There are a number of acceptable line
structures for CH3(CH2)4CH3, three of which are shown here. The orientation
of the line segments on the page is not important, but the number of line
segments connected to each point is important.
or

b.

CH2CH3

or

Bonds between carbon atoms and non-carbon atoms are indicated by line
segments that terminate with the symbol for the non-carbon atom. In this
case, the non-carbon atom is an oxygen.
O
or
O

c.

Bonds between hydrogens and non-carbon atoms are shown.

OH

or
OH

d.

The same rules apply for cyclic structures.

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12

Chapter 1

or

1.43

a.

7

b.

c.
H


H

O

H

H

H

H

H

C

C

C

C

C

C

C

H


H

H

H

H

H

C7H14O

H

1.44

First count the carbons, then the hydrogens, and finally the remaining atoms.
b.
C19H28O2
c.
C10H16
a.
C6H6
d.
C16H18N2O4S
e.
C9H6O2
f.
C10H14N2


1.45

a.

HONO

First determine the total number of valence electrons; H = 1,
O = 2 x 6 = 12, N = 5, for a total of 18. These must be
arranged in pairs so that the hydrogen has 2 and the other
atoms have 8 electrons around them.
H O N

or

O

H

O

N

O

Using the formula for formal charge given in Sec. 1.11, it can
be determined that none of the atoms has a formal charge.
b.

HONO2


There are 24 valence electrons. The nitrogen has a +1 formal
charge [5 – (0 + 4) = +1], and the singly bonded oxygen has a
–1 formal charge [6 – (6 + 1)] = –1. The whole molecule is
neutral.
O
H O N

(–1)
O

(+1)

c.

H2CO

There are no formal charges.
H
H C

d.

NH4+

O

The nitrogen has a +1 formal charge; see the answer to
Problem 1.25.
H
(+1)

H N H
H

e.

CN–

There are 10 valence electrons (C = 4, N = 5, plus 1 more
because of the negative charge).
–

C

N

The carbon has a –1 formal charge [4 – (2 + 3) = –1].
f.

CO

There are 10 valence electrons.
(–1)

C

O

(+1)

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Bonding and Isomerism

13

The carbon has a –1 formal charge, and the oxygen has a +1
formal charge. Carbon monoxide is isoelectronic with cyanide
ion but has no net charge (–1 + 1 = 0).
g.
BCl3 There are 24 valence electrons (B = 3, Cl = 7). The structure is
usually written with only 6 electrons around the boron.
Cl
Cl B
Cl

In this case, there are no formal charges. This structure shows
that BCl3 is a Lewis acid, which readily accepts an electron
pair to complete the octet around the boron.
H2O2

h.

There are 14 valence electrons and an O–O bond. There are
no formal charges.
H O O H

HCO3–


i.

There are 24 valence electrons involved. The hydrogen is
attached to an oxygen, not to carbon.
O
H O C O

O
(–1)

or

H

O

C

O

(–1)

The indicated oxygen has a formal charge: 6 – (6 + 1) = –1. All
other atoms are formally neutral.
1.46

This is a methyl carbocation, CH3+.
H
H


C

formal charge = 4 – (0 + 3) = +1

H

This is a methyl free radical, CH3.
H
H

C

formal charge = 4 – (1 + 3) = 0

H

This is a methyl carbanion, CH3– .
H
H

C

formal charge = 4 – (2 + 3 ) = –1

H

This is a methylene or carbene, CH2.
H

H


C

formal charge = 4 – (2 + 2) = 0

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14

Chapter 1

All of these fragments are extremely reactive. They may act as intermediates in
organic reactions.
1.47

There are 18 valence electrons (6 from each of the oxygens, 5 from the nitrogen, and
1 from the negative charge).
–1

O N

O

O

N O

or

–1

O

N

O

O

N

O

The negative charge in each contributor is on the singly bonded oxygen [6 – (6 + 1)
= –1. The other oxygen and the nitrogen have no formal charge. In the resonance
hybrid, the negative charge is spread equally over the two oxygens; the charge on
each is –1/2.

1.48

Each atom in both structures has a complete valence shell of electrons. There are no
formal charges in the first structure, but in the second structure the oxygen is
formally positive and the nitrogen formally negative.
H

O

CH3


C

H

C

C
H

1.49

C
H –1 H

H

In the first structure, there are no formal charges. In the second structure, the oxygen
is formally +1, and the ring carbon bearing the unshared electron pair is formally –1.
(Don’t forget to count the hydrogen that is attached to each ring carbon except the
one that is doubly bonded to oxygen.)
O

1.50

O

CH3

a.


O

H

CH3 C

OCH3

O

–1
OCH3
H

CH3

H

CH3 N

+1

b.

H

CH3

+1
O H


–1

O

C
CH3

1.51

+1
O CH3

O

CH3 C

O

OH

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Bonding and Isomerism

c.
1.52


15

S

CH3CH2

d.

H

CH3

O

CH2CH2

O

H

If the s and p orbitals were hybridized to sp3, two electrons would go into one of these
orbitals and one electron would go into each of the remaining three orbitals.
2p
sp 3
2s

The predicted geometry of ammonia would then be tetrahedral, with one hydrogen at
each of three corners, and the unshared pair at the fourth corner. In fact, ammonia has a
pyramidal shape, a somewhat flattened tetrahedron. The H–N–H bond angle is 107º.
1.53


The ammonium ion is, in fact, isoelectronic (the same arrangement of electrons) with
methane, and consequently has the same geometry. Four sp3 orbitals of nitrogen each
contain one electron. These orbitals then overlap with the 1s hydrogen orbitals, as in
Figure 1.9.

1.54

H

Cl

H

Cl

C

Cl

H

C

O

H

Cl


The geometry is tetrahedral at carbon. It does not matter whether we draw the wedge
bonds to the right or to the left of the carbon, or indeed “up” or “down”.
Cl

Cl C
Cl Cl

1.55

Cl

Cl

Cl

Cl

Cl
C

C

Cl

Cl

Cl

The bonding is exactly as in carbon tetrachloride. The geometry of silicon tetrachloride is
tetrahedral.

Cl
Cl Si
Cl
Cl

1.56

a.

OH

O

HO

H
H

H
H

H
H

H
A

b.

H H


OH

H H

H H
O H

HO

H

H

H

H H

H H

B

C

H

Structures A and B are identical. Structures A and C are isomers. They both
have the molecular formula C3H8O, but A is an alcohol and C is an ether.

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16

1.57

Chapter 1

Many correct answers are possible; a few are given here for part (a).
O

a.

C

H 3C

b.

OH
CH

CH 3

H2 C

O
H


C

C
H2

CH 2

CH 3

H2 C
HC

O

CH 3

CH 3

c.

CH

H2 C

O

O

H2 C


CH2

H 2C

1.58

Compounds c, e, h, and j all have hydroxyl groups (—OH) and belong to a class of
compounds called alcohols. Compounds b and k are ethers. They have a C—O—C
unit. Compounds f, i and I contain amino groups (—NH2) and belong to a family of
compounds called amines. Compounds a, d, and g are hydrocarbons.

1.59

The more common functional groups are listed in Table 1.6. Often more than one
answer may be possible.
a.

CH3CH2CH2OH
O

c.
H

e.

C

O

d.

C
H2

CH 3

H 3C

C

CH 3

O

CH3CH2 C

f.

CH3CH2OCH 2CH3

b.

OH

There are a number of possible answers. Five are shown below. Can you
think of more?
O
CH3CH2CH2CH2O

C


H

(CH3)2CHCH2O

O
CH3CH2CH2O

C

O

O
C

H

(CH3)3CO

C

H

O
CH3

CH3CH2O

C

CH2CH3


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Bonding and Isomerism

1.60

a.

17

carbonyl group (carboxylic acid), amino group (amine), aromatic group (arene)

b.

c.

O
HO

C

C9H11NO2

CHCH2
NH2

d.


The isomer of phenylalanine shown below is both an alcohol and an amide.
O

H2 N

C

CHCH2
OH

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2
Alkanes and Cycloalkanes; Conformational and
Geometric Isomerism
Chapter Summary
Hydrocarbons contain only carbon and hydrogen atoms. Alkanes are acyclic saturated
hydrocarbons (contain only single bonds); cycloalkanes are similar but have carbon rings.
Alkanes have the general molecular formula CnH2n+2. The first four members of this
homologous series are methane, ethane, propane, and butane; each member differs
from the next by a –CH2–, or methylene group. The IUPAC (International Union of Pure
and Applied Chemistry) nomenclature system is used worldwide to name organic
compounds. The IUPAC rules for naming alkanes are described in Secs. 2.3–2.5. Alkyl
groups, alkanes minus one hydrogen atom, are named similarly except that the -ane ending
is changed to -yl. The letter R stands for any alkyl group.
The two main natural sources of alkanes are natural gas and petroleum. Alkanes

are insoluble in and less dense than water. Their boiling points increase with molecular
weight and, for isomers, decrease with chain branching.
Conformations are different structures that are interconvertible by rotation about
single bonds. For ethane (and in general), the staggered conformation is more stable than
the eclipsed conformation (Figure 2.5).
The prefix cyclo- is used to name cycloalkanes. Cyclopropane is planar, but larger
carbon rings are puckered. Cyclohexane exists mainly in a chair conformation with all
bonds on adjacent carbons staggered. One bond on each carbon is axial (perpendicular to
the mean carbon plane); the other is equatorial (roughly in that plane). The conformations
can be interconverted by “flipping” the ring, which requires only bond rotation and occurs
rapidly at room temperature for cyclohexane. Ring substituents usually prefer the less
crowded, equatorial position.
Stereoisomers have the same order of atom attachments but different
arrangements of the atoms in space. Cis–trans isomerism is one kind of stereoisomerism.
For example, two substituents on a cycloalkane can be on either the same (cis) or opposite
(trans) sides of the mean ring plane. Stereoisomers can be divided into two groups,
conformational isomers (interconvertible by bond rotation) and configurational isomers
(not interconvertible by bond rotation). Cis–trans isomers belong to the latter class.
Alkanes are fuels; they burn in air if ignited. Complete combustion gives carbon dioxide and
water; less complete combustion gives carbon monoxide or other less oxidized forms of
carbon. Alkanes react with halogens (chlorine or bromine) in a reaction initiated by heat or
light. One or more hydrogens can be replaced by halogens. This substitution reaction
occurs by a free-radical chain mechanism.

18
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Alkanes and Cycloalkanes; Conformational and Geometric Isomerism


19

Reaction Summary
Combustion
Cn H2n +2

+

(3n2+1 )O

n CO2

2

+

(n + 1)H2 O

Halogenation (Substitution)
R

heat or light

H + X2

X + H

R


X

(X = Cl,Br)

Mechanism Summary
Halogenation
heat or light

Initiation

X

X

Propagation

R

H

+

X

R

+

H


X

R

+

X

X

R

X

+

X

Termination

2 X

2 X

X

X

2R


R

R

R

+

X

R

X

Learning Objectives
1.

Know the meaning of: saturated hydrocarbon, alkane, cycloalkane, homologous
series, methylene group.

2.

Know the meaning of: conformation, staggered, eclipsed, “dash-wedge” projection,
Newman projection, “sawhorse” projection, rotational isomers, rotamers.

3.

Know the meaning of: chair conformation of cyclohexane, equatorial, axial, geometric
or cis–trans isomerism, conformational and configurational isomerism.


4.

Know the meaning of: substitution reaction, halogenation, chlorination, bromination,
free-radical chain reaction, chain initiation, propagation, termination, combustion.

5.

Given the IUPAC name of an alkane or cycloalkane, or a halogen-substituted alkane
or cycloalkane, draw its structural formula.

6.

Given the structural formula of an alkane or cycloalkane or a halogenated derivative,
write the correct IUPAC name.

7.

Know the common names of the alkyl groups, cycloalkyl groups, methylene halides,
and haloforms.

8.

Tell whether two hydrogens in a particular structure are identical or different from one
another by determining whether they give the same or different products by
monosubstitution with some group X.

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